Chapter 9.8, Problem 3QQ

Interpretation Introduction

Interpretation:

From the given options, the change in position of the equilibrium with very small value for equilibrium constant has to be characterized.

Concept Introduction:

Law of Chemical Equilibrium:

The equilibrium constant is the product of molar concentrations of the product which is raised to its stoichiometric coefficients divided by the product of molar concentrations of the reactant which is raised to its stoichiometric coefficients.

Equilibrium Constant:

Consider a reaction,

    $\text{aA+bB}\iff \text{cC+dD}$

    Forward reaction rate ${\text{K}}_{\text{f}}$= ${\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{B}}$

    Backward reaction rate ${\text{K}}_{\text{b}}$= ${\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}$

    At equilibrium, the rate of forward reaction = rate of backward reaction

    $K_{\begin{array}{l}C\\ \end{array}}=\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{b}}}$ $\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{b}}}\text{=}\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{B}}}$

    ${\text{K}}_{\text{c}}\text{=}\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{b}}}\text{=}\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{B}}}$

${\text{K}}_{\text{C}}$ is the equilibrium constant.

If the value of the equilibrium constant is larger, then the equilibrium shifts to right.  That means the system have more products than reactants.

If the value of the equilibrium constant is smaller, then the equilibrium shifts to left.  That is the system will have more reactants than products.